$\dfrac{d}{dx}[7\text{cos}(x)+3e^x]=$
Explanation: Recall that ${\dfrac{d}{dx}[e^x]=e^x}$ and ${\dfrac{d}{dx}[\text{cos}(x)]=-\text{sin}(x)}$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[7\text{cos}(x)+3e^x] \\\\ &=7{\dfrac{d}{dx}[\text{cos}(x)]}+3{\dfrac{d}{dx}[e^x]} \\\\ &=7[{-\text{sin}(x)}]+3{e^x} \\\\ &=-7\text{sin}(x)+3e^x \end{aligned}$ In conclusion, $\dfrac{d}{dx}[7\text{cos}(x)+3e^x]=-7\text{sin}(x)+3e^x$